IIT-JEE 2002 Screening | Straight Lines and Pair of Straight Lines Question 17 | Mathematics

IIT-JEE 2002 Screening

MCQ (Single Correct Answer)

-0.75

Let $$P = \left( { - 1,\,0} \right),\,Q = \left( {0,\,0} \right)$$ and $$R = \left( {3,\,3\sqrt 3 } \right)$$ be three points.
Then the equation of the bisector of the angle $$PQR$$ is

$${{\sqrt 3 } \over 2}x + y = 0$$

$$x + \sqrt 3 y = 0$$

$$\sqrt 3 x + y = 0$$

$$x + {{\sqrt 3 } \over 2}y = 0$$

IIT-JEE 2002 Screening

MCQ (Single Correct Answer)

-0.75

A straight line through the origin $$O$$ meets the parallel lines $$4x+2y=9$$ and $$2x+y+6=0$$ at points $$P$$ and $$Q$$ respectively. Then the point $$O$$ divides the segemnt $$PQ$$ in the ratio

$$1 : 2$$

$$3 : 4$$

$$2 : 1$$

$$4 : 3$$

IIT-JEE 2002

MCQ (Single Correct Answer)

-1

A triangle with vertices $$(4, 0), (-1, -1), (3, 5)$$is

isosceles and right angled

isosceles but not right angled

right angled but not isosceles

neither right angled nor isosceles

IIT-JEE 2002

MCQ (Single Correct Answer)

-1

Locus of mid point of the portion between the axes of $$x$$ $$\cos \alpha + y\sin \alpha = p$$ where $$p$$ is constant is

$${x^2} + {y^2} = {4 \over {{p^2}}}\,\,\,$$

$${x^2} + {y^2} = 4{p^2}$$

$${1 \over {{x^2}}} + {1 \over {{y^2}}} = {2 \over {{p^2}}}$$

$${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}$$

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